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Raashan, Sylvia, and Ted play the following game. Each starts with $1. A bell rings every 15 seconds, at which time each of the players who currently have money simultaneously chooses one of the other two players independently and at random and gives $1 to that player. What is the probability that after the bell has rung 2019 times, each player will have $1? (For example, Raashan and Ted may each decide to give $1 to Sylvia, and Sylvia may decide to give her her dollar to Ted, at which point Raashan will have $0, Sylvia will have $2, and Ted will have $1, and that is the end of the first round of play. In the second round Rashaan has no money to give, but Sylvia and Ted might choose each other to give their $1 to, and the holdings will be the same at the end of the second round.)
A. 1/7
B. 1/4
C. 1/3
D. 1/2
E. 2/3
Are You Up For the Challenge: 700 Level Questions
A. 1/7
B. 1/4
C. 1/3
D. 1/2
E. 2/3
Are You Up For the Challenge: 700 Level Questions
12 Jun 2022, 22:12
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Raashan, Sylvia, and Ted play the following game. Each starts with $1. A bell rings every 15 seconds, at which time each of the players who currently have money simultaneously chooses one of the other two players independently and at random and gives $1 to that player. What is the probability that after the bell has rung 2019 times, each player will have $1? (For example, Raashan and Ted may each decide to give $1 to Sylvia, and Sylvia may decide to give her her dollar to Ted, at which point Raashan will have $0, Sylvia will have $2, and Ted will have $1, and that is the end of the first round of play. In the second round Rashaan has no money to give, but Sylvia and Ted might choose each other to give their $1 to, and the holdings will be the same at the end of the second round.)
A. 1/7
B. 1/4
C. 1/3
D. 1/2
E. 2/3
Are You Up For the Challenge: 700 Level Questions
A. 1/7
B. 1/4
C. 1/3
D. 1/2
E. 2/3
Are You Up For the Challenge: 700 Level Questions
It will never be the case that any of them is holding all $3 at the end of a given round. The explanation is that the most that a player can receive during a single round is $2, so in order to get to $3, they would need to already be holding the other $1, but if they're holding that $1 at the start of the round, they have to give it away, so the end the round with the $2 that were given to them. And if they only receive $1 during a round, that also means that they had to have started with $0 (can't get to 3), $1 (can't get to 3), or $2 (have to give one away, so can't get to 3).
So, we know that at the start of each round, the distribution is either 1-1-1 or 2-1-0.
Let's look at 1-1-1. Each player has 2 choices for who will receive their $1, so there are 2^3=8 possibilities of how the round concludes. Imagine all three players are sitting in a circle. The only way to get back to 1-1-1 is for everyone to pass to the left, or everyone pass to the right. So, there are 2/8 ways to stay at 1-1-1. 2/8 = 1/4.
Let's look at 2-1-0. Only two players pass a dollar. Each has 2 choices, so there are 2^2=4 possibilities of how the round concludes. The only way to get to 1-1-1 is for the person with 1 to give to the person with 0, and the person with 2 to give to the person with 1. So, there is 1/4 ways to get to 1-1-1 from 2-1-0. 1/4.
We've shown that it doesn't matter whether any given round starts 1-1-1 or 2-1-0, the probability of getting to 1-1-1 at the end of the round is 1/4.
Answer choice B.
If you have a bunch of time on your hands, want a break from GMAT, and are a nerd who likes stuff like this, go read about Markov Chains.
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